Homework Help: Momentum and Position Operator Commutator Levi Civita Form

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1. Nov 2, 2016

PatsyTy

1. The problem statement, all variables and given/known data
Prove that $[L_i,x_j]=i\hbar \epsilon_{ijk}x_k \quad (i, j, k = 1, 2, 3)$ where $L_1=L_x$, $L_2=L_y$ and $L_3=L_z$ and $x_1=x$, $x_2=y$ and $x_3=z$.

2. Relevant equations

There aren't any given except those in the problem, however I assume we use

$$L_i=x_jp_k-p_kx_j$$

$$[A,B]=AB-BA$$

$$[A+B,C]=[A,C]+[B,C]$$

$$[AB,C]=[A,C]B+A[B,C]$$

3. The attempt at a solution

Sub $L_i=x_jp_k-p_kx_j$ into $[L_i,x_j]$

$$[x_j-p_k-p_kx_j]=[x_jp_k,x_j]-[p_kx_j,x_j]$$

Use $[AB,C]=[A,C]B+A[B,C]$

$$([x_j,x_j]p_k+x_j[p_k,x_j])-([p_k,x_j]x_j+[x_j,x_j]p_k)$$

The two commutators $[x_j,x_j]=0$ so I end up with

$$x_j[p_k,x_j]-[p_k,x_j]x_j$$

However I have no idea where to go from here. I don't understand where the $x_k$ term comes from. I assume the $i\hbar$ term would appear if I wrote the momentum operators as $-i\hbar\frac{\partial}{\partial x_k}$ however then I would end up with a bunch of differential operators I don't know how to work with.

2. Nov 2, 2016

BvU

Easy: ${\partial \over \partial x_i }x_j = \delta_{ij}$

3. Nov 3, 2016

PeroK

This last equation is not correct. It is only true if the indices are $(i, j, k) = (1, 2, 3), (2, 3, 1)$ or $(3, 1, 2)$. For example:
$$L_1 = x_2 p_3 - x_3 p_2$$
This is where the Levi-Civita symbol comes in:
$$L_i = \epsilon_{ijk}x_j p_k$$
Because, if $i = 1$ say, then $\epsilon_{ijk} =0$ except $\epsilon_{123} =1$ and $\epsilon_{132} =-1$

And note that this invokes the "summation convention", which is a convenient shorthand and hides a whole bunch of the mathematical framework. For example, written out fully that would be:
$$For \ i = 1, 2, 3: \ \ L_i = \sum_{j,k = 1}^{3} \epsilon_{ijk}x_j p_k$$
That is three separate equations (one for each of $i = 1, 2, 3$) involving $L_i$ equal to a sum of 9 terms.

My advice is that, until you have really mastered this Levi-Civita symbol, you should always keep in mind that it is an extreme shorthand and often what appears to one equation involving a couple of terms is in fact much more complicated.

Last edited: Nov 3, 2016